Answer to Question #103122 in Statistics and Probability for Biraj Chhetri

Question

The standard deviation of a population is 2.70 cms. Find the probability that is random sample of size 66 (i) the sample mean will differ from the population mean by 0.75 cm or more, and (ii) the sample mean will exceed the population mean by 0.75 cm. or more (given that the value of the standard normal probability integral from 0 to 2.25 is 0.4877).

Answer 1

i) The probability that the sample mean will differ from population mean by 0.75 or more.

It implies that the probability that it exceeds +0.75cm or it is less than -0.75

P(-0.75 ≥$(\bar{x}-\mu)$ ≥0.75)

Convert to the standard normal using z=   (( $\bar{x}-\mu)*\sqrt{n}/\sigma$ )

=(±0.75)*$\sqrt{66}/2.7$

=±2.25

Given the value of the standard normal probability 0 to 2.25 is 0.4877

The probability that a value exceeds 2.25= 0.5-0.4877

=0.0123

The value is also the same for when a value is less than -2.25 since it is a standard normal

Hence the probability that the sample mean differs by 0.75 or more the population mean is

=0.0123+0.0123

=0.0246

ii) The sample mean will exceed the population mean by 0.75cm or more

P(($\bar{x}-\mu)$ ) ≥0.75)

Convert to the standard normal using z=  $(( \bar{x}-\mu)*\sqrt{n}/\sigma)$  

=$(0.75)*\sqrt{66}/2.7$

= =2.25

Given the value of the standard normal probability from 0 to 2.25 is 0.4877

The probability that a value exceeds 2.25= 0.5-0.4877

=0.0123